A jet airplane touches down on an airport runway with a speed of 203 km/hr. It...
A jet airplane touches down on an airport runway with a speed of 203 km/hr. It slows at a constant rate to a stop over a time period of 9.65 s. How far does the jet travel along the runway before stopping?
Homework Answers
Given initial velocity, u = 203 km/hr = 56.39 m/s
Final velocity, v = 0
t = 9.65 s
a = -a (because negative acceleration)
v = u -at
0 = 56.39-a*9.65
=> a = 5.84 m/s^2
Now, from 2nd eqn of motion,
s = ut +(1/2)at^2 = 56.39*9.65 - (1/2)(5.84)(9.65)^2 = 272.2 m (Answer)
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